Problem: The graph of a sinusoidal function intersects its midline at $(0,-2)$ and then has a minimum point at $\left(\dfrac{3\pi}{2},-7\right)$. Write the formula of the function, where $x$ is entered in radians. $f(x)=$
Solution: The strategy First, let's use the given information to determine the function's amplitude, midline, and period. Then, we should determine whether to use a sine or a cosine function, based on the point where $x=0$. Finally, we should determine the parameters of the function's formula by considering all the above. Determining the amplitude, midline, and period The midline intersection is at $y={-2}$, so this is the midline. The minimum point is $5$ units below the midline, so the amplitude is ${5}$. The minimum point is $\dfrac{3\pi}{2}$ units to the right of the midline intersection, so the period is $4\cdot \dfrac{3\pi}{2}={6\pi}$. [Why did we multiply by 4?] Determining the type of function to use Since the graph intersects its midline at $x=0$, we should use the sine function and not the cosine function. This means there's no horizontal shift, so the function is of the form $a\sin(bx)+d$. [How do we know that?] Determining the parameters in $a\sin(bx)+d$ Since the midline intersection at $x=0$ is followed by a minimum point, we know that $a<0$. [How do we know that?] The amplitude is ${5}$, so $|a|={5}$. Since $a<0$, we can conclude that $a=-5$. The midline is $y={-2}$, so $d=-2$. The period is ${6\pi}$, so $b=\dfrac{2\pi}{{6\pi}}=\dfrac{1}{3}$. The answer $f(x)=-5\sin\left(\dfrac{1}{3}x\right)-2$